# Q3(b) Fuzzy relations and composition [10]

Subject: Intelligent Systems II

Exam: TEE

Reader: https://deep-learning-study-room.vercel.app/is2/#tee/papers/paper-reexam-2026/q3b

Canonical: https://deep-learning-study-room.vercel.app/read/is2/tee/papers/paper-reexam-2026--q3b/

Actual TEE re-exam · 100 marks · 3 hours · Q1 compulsory + any 4 of Q2–Q7 · every answer verified (key errors flagged)

**Question. A = {0.6/small + 0.4/medium + 0.1/large}, B = {0.4/slow + 0.8/moderate + 0.3/fast}, C = {0.5/small + 0.2/medium + 0.4/large}. Find R = A × B, S = C × B and the max–min composition S ∘ R.** [10 marks]

### Answer

#### Given

The fuzzy sets or relation matrices and their listed row/column order.

#### To find

The relation matrix/matrices and the composition specified in the question.

#### Formula and why it applies

For Cartesian products take min of row and column membership. For the ordered matrix composition shown here:



$$
T_{ik}=\max_j\min(S_{ij},R_{jk})
$$



If max–product is asked instead, replace the inner min by multiplication.

#### Solution

##### Step 1. R and S



$$
R=\begin{bmatrix}0.4&0.6&0.3\\0.4&0.4&0.3\\0.1&0.1&0.1\end{bmatrix},\qquad S=\begin{bmatrix}0.4&0.5&0.3\\0.2&0.2&0.2\\0.4&0.4&0.3\end{bmatrix}
$$



##### Step 2. S ∘ R



$$
(1,1)=\max[\min(.4,.4),\min(.5,.4),\min(.3,.1)]=\max(.4,.4,.1)=0.4
$$





$$
(1,2)=\max[\min(.4,.6),\min(.5,.4),\min(.3,.1)]=\max(.4,.4,.1)=0.4
$$





$$
(1,3)=\max[\min(.4,.3),\min(.5,.3),\min(.3,.1)]=0.3
$$





$$
\text{row 2 (all 0.2)}=(0.2,0.2,0.2);\quad(3,1)=\max(.4,.4,.1)=0.4,\;(3,2)=\max(.4,.4,.1)=0.4,\;(3,3)=0.3
$$



**Final answer.** 

$$
R=\begin{bmatrix}.4&.6&.3\\.4&.4&.3\\.1&.1&.1\end{bmatrix}
$$

, 

$$
S=\begin{bmatrix}.4&.5&.3\\.2&.2&.2\\.4&.4&.3\end{bmatrix}
$$

, 

$$
S\circ R=\begin{bmatrix}0.4&0.4&0.3\\0.2&0.2&0.2\\0.4&0.4&0.3\end{bmatrix}
$$

. (The key gives R and S only.)

## Every cell without the row shortcuts

Both printed relations run from size to speed; their middle labels do not match for genuine relation composition. The numerical working below follows the question’s intended row-of-S with column-of-R operation using explicitly assumed positional alignment of the three middle entries.

### R: Cartesian-product cells

| Cell | Row degree | Column degree | min result |
| --- | --- | --- | --- |
| R11 | 0.6 | 0.4 | 0.4 |
| R12 | 0.6 | 0.8 | 0.6 |
| R13 | 0.6 | 0.3 | 0.3 |
| R21 | 0.4 | 0.4 | 0.4 |
| R22 | 0.4 | 0.8 | 0.4 |
| R23 | 0.4 | 0.3 | 0.3 |
| R31 | 0.1 | 0.4 | 0.1 |
| R32 | 0.1 | 0.8 | 0.1 |
| R33 | 0.1 | 0.3 | 0.1 |

### S: Cartesian-product cells

| Cell | Row degree | Column degree | min result |
| --- | --- | --- | --- |
| S11 | 0.5 | 0.4 | 0.4 |
| S12 | 0.5 | 0.8 | 0.5 |
| S13 | 0.5 | 0.3 | 0.3 |
| S21 | 0.2 | 0.4 | 0.2 |
| S22 | 0.2 | 0.8 | 0.2 |
| S23 | 0.2 | 0.3 | 0.2 |
| S31 | 0.4 | 0.4 | 0.4 |
| S32 | 0.4 | 0.8 | 0.4 |
| S33 | 0.4 | 0.3 | 0.3 |

**Result cell (1,1)**



$$
\max\{\min(0.4,0.4),\min(0.5,0.4),\min(0.3,0.1)\}
$$





$$
=\max\{0.4,0.4,0.1\}=0.4
$$



**Result cell (1,2)**



$$
\max\{\min(0.4,0.6),\min(0.5,0.4),\min(0.3,0.1)\}
$$





$$
=\max\{0.4,0.4,0.1\}=0.4
$$



**Result cell (1,3)**



$$
\max\{\min(0.4,0.3),\min(0.5,0.3),\min(0.3,0.1)\}
$$





$$
=\max\{0.3,0.3,0.1\}=0.3
$$



**Result cell (2,1)**



$$
\max\{\min(0.2,0.4),\min(0.2,0.4),\min(0.2,0.1)\}
$$





$$
=\max\{0.2,0.2,0.1\}=0.2
$$



**Result cell (2,2)**



$$
\max\{\min(0.2,0.6),\min(0.2,0.4),\min(0.2,0.1)\}
$$





$$
=\max\{0.2,0.2,0.1\}=0.2
$$



**Result cell (2,3)**



$$
\max\{\min(0.2,0.3),\min(0.2,0.3),\min(0.2,0.1)\}
$$





$$
=\max\{0.2,0.2,0.1\}=0.2
$$



**Result cell (3,1)**



$$
\max\{\min(0.4,0.4),\min(0.4,0.4),\min(0.3,0.1)\}
$$





$$
=\max\{0.4,0.4,0.1\}=0.4
$$



**Result cell (3,2)**



$$
\max\{\min(0.4,0.6),\min(0.4,0.4),\min(0.3,0.1)\}
$$





$$
=\max\{0.4,0.4,0.1\}=0.4
$$



**Result cell (3,3)**



$$
\max\{\min(0.4,0.3),\min(0.4,0.3),\min(0.3,0.1)\}
$$





$$
=\max\{0.3,0.3,0.1\}=0.3
$$